Physical Sciences Grade 12 Paper 1 Limpopo June 2025 Memo

PROVINCIAL GOVERNMENT REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE Pe 4 f UY in a a i it i it s J PHYSICAL SCIENCES: PHYSICS (P1) JUNE 2025 FINAL AND AMENDE MARKING GUIDELINE/ FINALE EN AANGEPASTE MERK RIGLYNE These marking guidelines consist of 9 pages POPPER UPECTORTETORCTEOTTLECTITTTT TTT kd MARKS: 150

Limpopo DoE/June 2025 Copyright reserved ; please turn over QUESTION 1 11. A¥¥ 12 Cvv 13. A¥vv 14 Dv¥v 15 BYY 16 BYY 17 Cv¥v 18 Bvv 19 Avv 1.10 Dvv (2 x 10 =20) QUESTION/VRAAG 2 2.1 Abody will remain in its state of rest or motion at constant velocity unless a 2:2 non-zero resultant/net force acts on it. ¥W ‘n Voorwerp sal in ‘n toestand van rus of konstante bewegin bly tot ‘n nie-sero resultante krag daarop inwerk. VV (2) Fa < fx Copyright reserved Please turn over

Limpopo DoE/June 2025 ACCEPTED SYMBOLS/AANVAARDE SIMBOLE e Mark awarded for label and arrow ¢ Do not penalize for length of arrows e Any additional force/s: -1 e If force(s) do not make contact with the body: -1 mark e -1 mark if all arrowheads are omitted but correctly labelled. FN | Normal force/N /Fesurface on box Normaalkrag/ N/Fopperviak op krat fk : Kinetic friction / friction /Frictional | Kinetiese | force wrywing/wrywing/wrywingskrag Fg "w, weight/Gravitational force/ Gewing/Gravitasiekrag/ ;_ Fearth on box Faarde op die voorwerp Fa/F Ap F7/ FToegepaste t ‘2c ars Notes: Notas: e Punt toegeken vir pyl en pylpunt e Moenie lengte van pyle penaliseer nie e Addisionele krag/te:-1 e As kragte nie kontak maak met kol nie: -1 e -1 punt as alle kragte teenwoordig (4) is en benoem maar geen pylpunte nie. 2.3.1 OPTION 1 NOTES: Free = 0 7] Formula for N1/N2 Fret = ma ¥ Any one/ Enige een Substitution of zero Fax — f,, — mgsiné = 0 3f,cos28°Y — f, — (48 + 4,3)(9,8)sin15° Vv = 0 ¥ f, = 80,45 NV Substitution for Fax Substitution for Fgu Answer = 80,45 N {This value is checked and correct) OPTION 2 fy = Np ¥ = ux(mgcos@; — Fr sin @2) = 0,22v (48 +4,3)(9,8)cos15 v — 3fsin28v’) NOTES: Formula for fx Substitution of 0,22 Substitution for Fax Substitution for Fgu = 83,15N v Copyright reserved Answer = 83,15 N Please turn over

Limpopo DoE/June 2025 2.3.2 POSITIVE MARKING FROM QUESTION 2.3.1 Tension in the rope is the applied force/ Spanning in die tou is die toegepaste krag Fy, = 3f, ¥ (Ratio) Fa = 3¥(80,45)¥ Fa= 241,35 NY OR Range: 241,35 — 249,46 N OPTION 2 Fret = Fy -— Fey —-i¥ OR 0 = Facos 28 — 52,3(9,8)sin 15“ — 80,45 ¥ Fa = 241,36 NY OPTION 3 Fx2=Fy2+ Fy? v = (3fc0s28)? + (3fsin28)?_¥ = (3(80,45)cos28)? + (3(80,45)sin28)? v = (213,21)? + (113,36) Fa=241,47N Vv F, = 3f, ¥ (Ratio) Fa = 3¥(83,15)¥ Fa =249,4557 Fa= 249,46 NV Faae= Feo Foy — fy ¥ 0 = Facos 28 — 52,3(9,8)sin 15” — 83,15 ¥ Fa = 244,41 NY (4) 2.3.3 (POSITIVE MARKING FROM QUESTION 2.3.1 and 2.3.2 OPTION 1 fi, = Ni 80,45 = N (0,22) ¥ N = 365,68 NV 377,95 N OR Copyright reserved fy = Ni 83,15 = N (0,22) v N = 377,95NY Please turn over

OPTION 2: Fret = 0 N = Fg. - Fy v Any One = mg cos81 — Fa sin@2 = (48 + 4,3)(9,8)(cos15) — 241,35 sin 28 Vv = 381,76866 = 381,77NV Limpopo DoE/June 2025 (3) 2.4 Rate at which work is done/ Rate at which energy is expended/transferred VV Tempo waarteen arbeid verrig word/ Tempo waarteen energie oorgedra word. v~~~ (2 OR 0) (2) 2.5 (POSITIVE MARKING FROM QUESTION 2.3.2) check alternative answers from 2.3.2 Ww = FA x cos@ calculate Fx 241,35(5)(cos28°) ¥ OR Wex = Fx Axcos@ = 1065,497 J Range: 1065,497 - 1101,43 J = (241,35c0s28)(5)(cos0) ¥ = 1065,497 J w Carry over mark. Correctly P= t v v substituting the value calculated for work in the power formula. But then 1065,497 ‘ P = ——— the final (wrong) answer is not aN awarded a mark. = 332,9678 = 332,97 W¥ Range: 332,97- 344,20 W (5) Copyright reserved Please turn over

Limpopo DoE/June 2025 OPTION 3: Speed = — = — = 1,5625 ms* at 3,2v Carry over mark. Correctly substituting speed in power formula Pave =FaxVave ¥ (Fa fropé 2.3.2 Range: 241,35 — 249,49 N = (241,35c0s 28) ¥ (1,5625) ¥ = 332,97 W v (Range: 332,97 - 344,20W) QUESTION 3 | In an isolated system, the total linear momentum is conserved/ remains constant. “~~ In ‘n geisoleerde sisteem sai die totale lineére momentum behoue bly/constant bly. WW (2 or 0) (2) B22) LP = Lpr Y Anyone/ Enige een my, = ny + Mp (90000 + mpr)4500Y = (90000)(11000)¥ + m,-(—3045)¥ Mer = 77 534,79 kg¥ OPTION 2 Apy = —Appr ¥ m(vp — Yj) = —M(vy — Vj) 9000 (11000V — 4500v) = -m(4500- (-3045) Vv m=7753,48kg ¥ (5) 3.3 POSITIVE MARKING FROM 3.2 figs = mv? ise vAny one/ Enige een = + (90000 + 77534,79)(4500)? A = 1,696 x 1014] 1 2 1 2 Exe = zw + gm = = (90000)(11000)? +5 (77534,79)(3045)°¥ =5,8 x 1077] Bay # Exe Decoupling is not elastic“ / Ontkoppeling is onelasties (5) Copyright reserved Please turn over

Limpopo DoE/June 2025 QUESTION 4 44 3 m:s1 downwardsv /afwaarts (1) 42 N IF § 2 vp = vj + 2ady 1 2 = (-3)? + 2(-9,8)(-18) Ayah + pat , = —19,02 m-s7 ve anil -18 V = (-3)t +% (-9,8)t2 ve= vyitadtv t=1,63s” 19,02V = 3+ 9,8Atv eons At = 1,63 s¥ (4) 4.3.1 For the follov.:ng, options, vandidate calculated final velocity first: vp = vp + 20d = (+37 +26 vi = -19,02104 =? Frec(tA) = mvp — v,)~ Ay = (2) ty rag. At = (vr -vi) —19,02104+(-3 -9,8 (At) = (-19,02104 — (-3)) ¥ -18 = eae yat ee At = 1,63 sv Calculate the distance that the crane moves in 1 second at constant velocity: Bereken die afstand wat die hyskraan in 1 sekonde beweeg teen konstante snelheid: Ay = viAt+Fa(At)? ¥ = 3(1) V +0 (non-essential zero may be omitted) Ay = 3m distance vitve =———__ ¥ Ay = ( ) tv sbaed time y 2 i = i =—3-3 Distance = speed x time _ ( - ) ()v = 3(1)¥ =3m = 3m C was thrown from: 18 — 3 = 15 mv (3) Copyright reserved Please turn over

Limpopo DoE/June 2025 8 4.3.2 POSITIVE MARKING 4.2 and 4.3.1 IF OPTION 1: Ay = v,At+ 5 a(At)? v ~15 = v;(0,63) +5 (9,8) (0,63)? vy; = —20,72 ms" vy, = 20,72 m.s‘! Ydownwards ¥ Range: 20,519 — 20,72 OPTION 2: Use of energy principles: Emech top = Emech bottom Ep + Extop = Ep + Exbottom Any one ¥ mgh een? = 0=mv? 2 2 (9,8)(15) + Ye v2 = % (26,74)°v vit = 20,519 m.s! ¥ downward ¥ (4) 4.3.3. POSITIVE MARKING FROM QUESTION 4.2 Crane moves at constant velocity for 1,63s./ Hyskraan beweeg teen konstante snelheid vir 1,63s. Ay = vAt¥ Ay = (=) tv = (3)(1,63) ¥ OR eer = 4,89 m = (=) (1,63) v = 4,89 | Height = 18 =“ 4,89 (Mark is for subtracting the answer from 18) = 13,114mv (3) Copyright reserved Please turn over

Limpopo DoE/June 2025 9 44 POSITIVE MARKING FROM 4.2, 4.3.1, 4.3.2, 4.3.3 18 4 13,11 = c co] £ Time {s) 1 163 NOTES: : NOTAS: Starting height (AB) ¥ Drop height of C ¥ Throw time of C ¥ Fina position of A“ B &C hit ground together v Final time for atl graphs “ Aanvanklike hoogte (AB) “ Gooi/val hoogte van C ~ Gooi tyd van C ~ Finale posisie van A ~ B & C tref grond op dieselfde tyd “ Finale tyd vir al drie grafieke ~ (6) QUESTION 5 5.1 A system on which the external net force is zero. VW ‘n Sisteem waarop die eksterne netto krag nulis VV (2 or 0) (2) 5.2 * Any additional forces/Addisionele kragte: -1 | * Allow w/ weight /Aanvaar gewig Fov (1) 5.3 Mechanical Energy gottom = Mechanical Energyrop Meganiese energieonder = Meganiese energiebo ty Ex) bottom = (Ep 5 ee Any one ¥ (mgh += mv?) = (mgh +smv*) 2 bottom 2 top Wie — Alig + ABR NOTES: Wet = BE, Y Formula 0 +5 (S2)(v*)” = (52)(9,8)(0,47) ¥+ 0 V Subs Ep vy, =3,04m-s tv v Subs Ek v Answer (4) Copyright reserved Please turn over

5.4 Ap = mv; — mv; Ap = m(vr— vi) = 52(3,04 — 0)¥ = 158,08 kg: m-s~! V upwards¥ /opwaarts 10 iY 5.5 POSITIVE MARKING FROM 5.4 158.08 N-s upwardsY /Opwaarts 5.6 EreAt= Ap | OR/OF Ap Fret = At OR/OF Fnet & x AND: momentum remains constant “ OR: Any onev” Limpopo DoE/June 2025 (4) (1) If time increase and force decreaseY,, the momentum will remain the same” As tyd toeneem en krag afneem V vir dieselfde verandering in momentum, V Copyright reserved (2) [14] Please turn over

Limpopo DoE/June 2025 11 QUESTION 6 6.1 A force for which work done in moving an object between two points depends on the path taken. vw ‘n krag waarvoor die arbeid verrig word om die voorwerp tussen twee punte beweeg afhang van die pad van beweging.V¥ (2 or 0) (2) 6.2 W, = fAx cosO¥ = (7,5)(5)cos180°Y = -37,5]v (3) 6.3. POSITIVE MARKING FROM 6.2 NOTE: Wac — AE, + AE, - Wnet = AE v Subst Ws Weg Cd We = AEx v Subst AEp v Subst AEx v Final answer 1 Wrc = (mghs — mgh,) + 2 m(v7 = vi) -37,5Y = (35)(9,8)(20 x 10% - 2,5) ¥ + % (35)(v? — 0,37) ¥ Vp = 656m:s tv NOTE: If equations of motion is used: 0/5 (5) [10] QUESTION 7 7.1 Doppler effect ¥ Change in the frequency/pitch of the sound detected by a listener, because the sound source and the listener have different velocities relative to the medium of sound propagation. vv / Doppler effek Verandering in frekwensie/toonhoogte van die waargenome klank agv relatiewe beweging tussen die bron en luisteraar (3) 7.2.1. Speed (of car)/Motion (of car) /Spoed (van die motor) Vv (1) 7.2.2 (observed) frequency” / (waargenome)frekwensie (1) oe BY (1) 74 The detected frequency is higher “/Die waargenome frekwensie is hoér (1) Copyright reserved Please turn over

Limpopo DoE/June 2025 75 cee Vv =) = ae v fi ( V+Vs V;=55,24m:-s"1¥ (4) [11] QUESTION 8 8.1 The magnitude of the electrostatic force exerted by one point charge (Q1) on another point charge (Q2) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them:¥v/ Twee ladings sal ‘n elektrostatiese krag uitoefen deur een puntlading op ‘n ander is direk eweredig aan die produk van die ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle. (2) 8.2 Fe/Fc/Faons ¥ (Coulombic force) Fov (2) 8.3 Fe a v MARKING CRITERIA: re 2 , (20 x 10-3)(9,85 2 — -v ¢ Coulomb’s Law (25xa0%) ¢ Correct substitution into Fg and Fe Q=1,17 x 10-7C¥ : e Equating Fg and Fe e Final answer (sign not considered) (4) Copyright reserved Please turn over

Limpopo DoE/June 2025 13 8.4 \ * All lines curve away (shape) /alle lyne buig weg (vorm) . SAIL i * All arrows point towards the charge (direction ; _ 3 ‘ of the field lines)V/Alle pyle wys na die i Ye lading(rigtin van veldlyne) a4 * No lines touching/crossing and all make contact with the chargesY /Geen lyne —\ Loa raak/kruis of maak kontak met die ladings nie a (3) 85 pK < v _ (9 x 10°)(1,17 x 107-7)¥ - (25 x 10-3)2v = 1,68 x 10°N:Cl¥ (4) QUESTION 9 9.1 The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature. VW (2 or 0) Die potensiaalverskil oor die punte van ‘n geleier is direk eweredig aan die stroom in die geleier mits temperatuur constant bly (2) 9.2.1 Vinternal = Ir 7 L7=1(0,7)¥ I = 2,43 AY (3) 9.2.2 (POSITIVE MARKING FROM QUESTION 9.2.1) Ra: V = IRY 5 = 2,43 RY Rz = 2,060 Ri =2RoV¥ :.Ri=4,120 Vv (4) 9.2.3 (POSITIVE MARKING FROM QUESTIONS 9.2.1 AND 9.2.2) OPTION 1 e=I(R+rn)¥ = (2,43)(0,7 + 2,06+ 4,12) Vv € = 16,72 WW Copyright reserved Please turn over

9.3.1 14 OPTION 2 lar = Ire Var _ V2 Ry Re 4% S8y 2k, Rz Vi=10V €=1,7+10+5V =16,7VV (POSITIVE MARKING FROM 9.2) Rp = 1,811 Rr = Rp +R, Ry = 1,811 + 4,12V Ry = 5,930. e=I(R+nVv 16,72 = 1(5,9 0.7)” 1=2,52AV Copyright reserved Limpopo DoE/June 2025 (3) MARKING CRITERIA e Correct substitution into R,/Korrekte invervanging e Addition of Rp and R1/Som van Rp en Ri e EMF formula/ EMK formule e Substituting into EMF formula Invervanging in EMK formule e Final answer/Finale antwoord (5) Please turn over

9.3.2 15 Limpopo DoE/June 2025 (POSITIVE MARKING FROM QUESTION 9.3.1) MARKING CRITERIA V over Ra: ¢ Substituting into Ohm's V=IR law/Invervanging in Ohm se _ wet = 2,52(1.811) ¥ =456 e Power formula/Drywing formule e Substituting into power P= ve Y formula//nvervanging in gece drywing P= ; ne v e Final answer/Finale antwoord P = 10,09 WY OPTION 2 lin Re 1=2,52Y x 17,02 1=2,221A R=lRY = (2,221)?(2,52) ¥ = 10,16 WY OPTION 3 lin Re 15 1=2,52Y x 17,02 1=2,221A P=VIV¥ = (4,56)(2,221) ¥ = 10,127WV¥ RANGE: 10,09 — 10,16 W (4) [21] GRAND TOTAL/GROOTTOTAAL: 150 Copyright reserved Please turn over